Examine where $\nabla f = (0, 0)$

Question

Given $f: \Bbb R^2 \rightarrow \Bbb R$ defined by

$$f(x) = \|x\|^4 - \|x\|^2$$

where $x = (x_1, x_2) \in \Bbb R^2$, I have to examine where $\nabla f = (0, 0)$.

Approach

Since $||x||$ was not defined, I can only assume that this is the euclidian norm, which means that,

$||x||^4 := (x_1)^2 + (x_2)^2,$

$||x||^2 := x_1 + x_2.$

Therefore,

$df \over dx_1$ = $2x_1,$ $df \over dx_2$ =$ 2x_2,$

such that,

$\nabla f = (2x_1, 2x_2) = (0, 0)$

only holds for $(x_1, x_2) = (0, 0)$.

I have to admit that this seems a little bit too easy. Did I make a mistake when I defined $||x||^4$ and $||x||^2$?

Answer 1

First we have : $$ f(x) = ((x_1)^2 +(x_2)^2)^2 - (x_1^2 + x_2^2) = x_1^4 + 2x_1^2x_2^2 + x_2^4 -x_1^2 -x_2^2 $$ Then we have $$ \frac{\partial}{\partial x_1} f(x) = 4x_1^3 + 4x_1 x_2^2 - 2x_1 = 0 $$ $$ \frac{\partial}{\partial x_2} f(x) = 4x_2^3 + 4x_1^2 x_2 - 2x_2 = 0 $$

Clearly $(0,0)$ is a solution. We can then divide the first equation by $x_1$ and the second by $x_2$ : $$ 4x_1^2 + 4x_2^2 - 2 = 0\\ 4x_2^2 + 4x_1^1 - 2 = 0 $$ We see that both equations are actually equivalent : $x_1^2 + x_2^2 = 1/2$ which is the equation of the circle centered at $(0,0)$ of radius $\sqrt{1/2}$.

Therefore the gradient is $0$ at the origin and on this circle.

Answer 2

Assuming that $\|\cdot\|$ denotes the $2$-norm,

$$f (x) = \|x\|_2^4 - \|x\|_2^2 = \|x\|_2^2 \, (\|x\|_2^2 - 1) = (x^T x) \cdot (x^T x - 1)$$

Taking the gradient,

$$\nabla f (x) = 2 (\|x\|_2^2 - 1) \, x + 2 \|x\|_2^2 \, x = 2 \, (2 \|x\|_2^2 - 1) \, x$$

Hence, the gradient vanishes at the origin, $x = 0_2$, and on the circle $\|x\|_2 = \frac{1}{\sqrt 2}$.